YES
We show the termination of the TRS R:

  le(|0|(),y) -> true()
  le(s(x),|0|()) -> false()
  le(s(x),s(y)) -> le(x,y)
  minus(x,|0|()) -> x
  minus(s(x),s(y)) -> minus(x,y)
  mod(|0|(),y) -> |0|()
  mod(s(x),|0|()) -> |0|()
  mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
  if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
  if_mod(false(),s(x),s(y)) -> s(x)


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: le#(s(x),s(y)) -> le#(x,y)
p2: minus#(s(x),s(y)) -> minus#(x,y)
p3: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))
p4: mod#(s(x),s(y)) -> le#(y,x)
p5: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p6: if_mod#(true(),s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: if_mod(false(),s(x),s(y)) -> s(x)

The estimated dependency graph contains the following SCCs:

  {p3, p5}
  {p1}
  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: if_mod#(true(),s(x),s(y)) -> mod#(minus(x,y),s(y))
p2: mod#(s(x),s(y)) -> if_mod#(le(y,x),s(x),s(y))

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1, r2, r3, r4, r5

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        if_mod#_A(x1,x2,x3) = max{6, x2 - 23}
        true_A = 4
        s_A(x1) = max{20, x1 + 12}
        mod#_A(x1,x2) = max{5, x1 - 13}
        minus_A(x1,x2) = max{3, x1 + 1}
        le_A(x1,x2) = max{x1 + 1, x2 + 3}
        |0|_A = 20
        false_A = 11
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          if_mod#_A(x1,x2,x3) = max{2, x1 - 19, x2 - 56, x3 - 45}
          true_A = 9
          s_A(x1) = x1 + 46
          mod#_A(x1,x2) = max{2, x1 - 56, x2 - 45}
          minus_A(x1,x2) = max{x1 + 40, x2 + 34}
          le_A(x1,x2) = max{x1 - 9, x2 + 9}
          |0|_A = 10
          false_A = 20
          if_mod#!_A(x1,x2,x3) = max{1, x1 - 6, x2 - 93, x3 - 43}
          true!_A = 13
          s!_A(x1) = x1 + 4
          mod#!_A(x1,x2) = 2
          minus!_A(x1,x2) = max{5, x1 - 47}
          le!_A(x1,x2) = max{58, x2 + 12}
          |0|!_A = 0
          false!_A = 58
  
    partial status:
  
      pi(if_mod#) = []
      pi(true) = []
      pi(s) = [1]
      pi(mod#) = []
      pi(minus) = []
      pi(le) = [1]
      pi(|0|) = []
      pi(false) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: le#(s(x),s(y)) -> le#(x,y)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        le#_A(x1,x2) = 1
        s_A(x1) = 2
        le_A(x1,x2) = max{5, x1 - 4, x2 - 4}
        |0|_A = 1
        true_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          le#_A(x1,x2) = max{x1, x2}
          s_A(x1) = max{6, x1 + 1}
          le_A(x1,x2) = max{2, x1 - 7, x2 - 7}
          |0|_A = 6
          true_A = 0
          le#!_A(x1,x2) = max{3, x1 - 2, x2 - 7}
          s!_A(x1) = 2
          le!_A(x1,x2) = max{5, x1 - 14, x2 - 7}
          |0|!_A = 1
          true!_A = 3
  
    partial status:
  
      pi(le#) = []
      pi(s) = []
      pi(le) = []
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: minus#(s(x),s(y)) -> minus#(x,y)

and R consists of:

r1: le(|0|(),y) -> true()
r2: le(s(x),|0|()) -> false()
r3: le(s(x),s(y)) -> le(x,y)
r4: minus(x,|0|()) -> x
r5: minus(s(x),s(y)) -> minus(x,y)
r6: mod(|0|(),y) -> |0|()
r7: mod(s(x),|0|()) -> |0|()
r8: mod(s(x),s(y)) -> if_mod(le(y,x),s(x),s(y))
r9: if_mod(true(),s(x),s(y)) -> mod(minus(x,y),s(y))
r10: if_mod(false(),s(x),s(y)) -> s(x)

The set of usable rules consists of

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        minus#_A(x1,x2) = max{7, x1 - 7, x2 - 16}
        s_A(x1) = max{15, x1 + 6}
        le_A(x1,x2) = max{4, x1 - 16, x2 - 16}
        |0|_A = 2
        true_A = 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          minus#_A(x1,x2) = max{x1, x2}
          s_A(x1) = max{3, x1 + 1}
          le_A(x1,x2) = max{2, x1 - 4, x2 - 3}
          |0|_A = 2
          true_A = 0
          minus#!_A(x1,x2) = max{8, x1 - 4, x2 - 20}
          s!_A(x1) = max{7, x1 - 19}
          le!_A(x1,x2) = max{4, x1 - 18, x2 - 10}
          |0|!_A = 2
          true!_A = 3
  
    partial status:
  
      pi(minus#) = []
      pi(s) = [1]
      pi(le) = []
      pi(|0|) = []
      pi(true) = []

The following rules are removed:

  p1

Then no dependency pair remains.